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Is the series, $$1 - 24\sum_{n = 1}^\infty \frac{q^{2n}}{(1 - q^{2n})^2}, \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ somehow related to $$E_2(q) = 1 - 24\sum_{n = 1}^\infty \frac{nq^{2n}}{1 - q^{2n}}, \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ the Eisenstein series of weight 2?

glebovg
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1 Answers1

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It appears that they are one and the same. In fact, it is very easy to establish.

Let $\sum$ denote the summation that runs over all the natural numbers. Then \begin{align*} \sum_n \frac{q^{2n}}{(1 - q^{2n})^2} &= \sum_n \sum_m q^{2n} m q^{2n(m - 1)} \\ &= \sum_m \sum_n m q^{2nm} \\ &= \sum_m \frac{m q^{2m}}{1 - q^{2m}}. \end{align*}

glebovg
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    If you happen to enjoy such rearrangements, you can work out $$\begin{align} \operatorname{E}4 &= 1 + 240\sum{n=1}^\infty \frac{q^n}{(1-q^n)^2} \left(1 + 6\frac{q^n}{(1-q^n)^2}\right) \ \operatorname{E}6 &= 1 - 504\sum{n=1}^\infty \frac{q^n}{(1-q^n)^2}\left( 1 + 30\frac{q^n}{(1-q^n)^2}\left( 1 + 4\frac{q^n}{(1-q^n)^2}\right)\right) \end{align}$$ using $q = \exp(2\pi\mathrm{i}\tau)$ here. – ccorn Sep 24 '14 at 00:01
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    There is a general formula for $E_{2k}$ that sums polynomials of $q^n/(1-q^n)^2$. It involves central factorial numbers of the second kind with even index (OEIS A036969). As a consequence, series summing polynomials of $q^n/(1-q^n)^2$ can be represented as linear combinations of Eisenstein series. Besides, those rearranged series converge also for $|q|>1$ (but not for $|q|=1$, therefore they do not provide an analytic continuation) and yield the same result as with $1/q$ instead of $q$. – ccorn Sep 24 '14 at 00:21